Problem: Isaac's favorite cereal is running a promotion that says $1$ -in- $4$ boxes of the cereal contains a prize. Suppose that Isaac is going to buy $6$ boxes of this cereal, and let $X$ represent the number of prizes he wins in these boxes. Assume that these boxes represent a random sample, and assume that prizes are independent between boxes. What is the probability that he wins exactly $2$ prizes in the $6$ boxes? You may round your answer to the nearest hundredth. $P(X=2)=$
Explanation: Without a fancy calculator Winning $2$ prizes in $6$ boxes means Isaac needs to win a prize in $2$ boxes and not win a prize in $4$ boxes. Since $1$ -in- $4$ boxes contain a prize, we know $P({\text{win}})={25\%}$ and $P({\text{not}})={75\%}$. Since we are assuming independence, let's multiply probabilities to find the probability of winning prizes in $2$ boxes followed by not winning prizes in $4$ boxes: $P({\text{WW}}{\text{NNNN}})=({0.25})^2({0.75})^4\approx0.01978$ This isn't our final answer, because there are other ways to win $2$ prizes from $6$ boxes (for example, NNNNWW). How many different ways are there? We can use the combination formula to find how many ways there are to win $2$ prizes from $6$ boxes: $\begin{aligned} _n\text{C}_k&=\dfrac{n!}{(n-k)!\cdot k!} \\\\ _6\text{C}_2&=\dfrac{6!}{(6-2)!\cdot2!} \\\\ &=\dfrac{6 \cdot 5 \cdot \cancel{4 \cdot 3 \cdot 2 \cdot 1}}{(\cancel{4 \cdot 3 \cdot 2 \cdot 1}) \cdot 2 \cdot 1} \\\\ &=15 \end{aligned}$ There are $15$ ways to win prizes in $2$ of $6$ boxes. Do they all have the same probability? Each of the $15$ ways has the same probability that we already found: $\begin{aligned} P({\text{WW}}{\text{NNNN}})&=({0.25})^2({0.75})^4\approx0.01978 \\\\ P({\text{W}}{\text{N}}{\text{W}}{\text{NNN}})&=({0.25})^2({0.75})^4\approx0.01978 \\\\ &\vdots \\\\ P({\text{NNNN}}{\text{WW}})&=({0.25})^2({0.75})^4\approx0.01978 \end{aligned}$ So we can multiply this probability by $15$ since that is how many ways there are to win prizes in $2$ of $6$ boxes: $\begin{aligned} P(X=2)&=15(0.25)^2(0.75)^4 \\\\ &\approx15(0.01978) \\\\ &\approx0.29663 \\\\ &\approx0.30 \end{aligned}$ Answer $P(X=2)\approx0.29663\approx0.30$